Problem: $f(x)=6-\dfrac{1}{2}x$ $h(x)=4(x-3)^2$ Write $f(h(x))$ as an expression in terms of $x$. $f(h(x))=$
Answer: Let's write $h(x)$ as the input to function $f$. $ f({h(x)})=6-\dfrac{1}{2}\left({h(x)}\right)$ Since $h(x)=4(x-3)^2$, this becomes: $\begin{aligned} f({h(x)})&=6-\dfrac{1}{2}\left({4(x-3)^2}\right)\\ \\ &=6-2(x-3)^2\\ \\ &=6-2(x^2-6x+9)\\ \\ &=6-2x^2+12x-18\\ \\ &=-2x^2+12x-12 \end{aligned}$ Note: We simplified the result to obtain a nicer expression, but this is not necessary. The answer: $f(h(x))=-2x^2+12x-12$